Question

Which of the following statements is/are correct about the ionic product of water?

• A. K (equilibrium constant of water) < K${}_w$
• B. pK>pK${}_w$
• C. At ${100}^{0}C$ $Kw=7.5\times {10}^{-14}$ $(moles/lit{)}^2$
• D. Ionic product of water at ${25}^{0}Cis{10}^{-14}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A pK>pK${}_w$
B K (equilibrium constant of water) < K${}_w$
C Ionic product of water at ${25}^{0}Cis{10}^{-14}$
D At ${100}^{0}C$ $Kw=7.5\times {10}^{-14}$ $(moles/lit{)}^2$

$H_{2}O+H_{2}O\leftrightharpoons H_3{O}^{+}+O{H}^{-}$
This equilibrium applies to pure water and any aqueous solution.
Approximating activities by concentrations, the chemical equilibrium constant, Keq, for this reaction is given by:$K_{eq}=\frac{\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]}{[H_{2}O{]}^2}$
$K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]=K_{eq}\cdot [H_{2}O{]}^2$ So $K_{eq}<K_w$

At ${25}^0$ C Kw is equal to $1.0\times {10}^{-14}.$ and as temperature increases its value increases. At ${100}^{0}C$ its value becomes $7.5\times {10}^{-14}$
$pK=-lo{g}_{10}{K}_w$.

$\therefore$ $pK>p{K}_w$